A338610 Integers m such that there exist one prime p and one positive integer k, for which the expression k^3 + k^2*p is a perfect cube m^3.

2, 12, 36, 80, 252, 810, 1100, 1452, 2366, 2940, 5202, 12696, 14400, 16250, 20412, 22736, 27900, 33792, 40460, 52022, 56316, 70602, 75852, 93150, 112896, 120050, 143312, 169400, 198476, 242172, 254016, 291852, 305252, 410700, 518400, 538002, 643452, 689216, 737100

Offset: 1

Author

Bernard Schott, Nov 03 2020

Keywords

• nonn

Comments

This concerns Problem 131 of Project Euler (see link).

For each such term m with this property, the values of k and of p are unique.

The solution to the Diophantine equation is: (q^3)^3 + (q^3)^2 * ((q+1)^3 - q^3) = ((q+1) * q^2)^3, where

- the prime p is the cuban prime (q+1)^3 - q^3 = A002407(n),

- corresponding to q = A111251(n),

- the positive integer k = q^3, and,

- the resulting m = (q+1)*q^2 = (A111251(n)+1)*(A111251(n))^2.

Examples

			For n=1, q=A111251(1)=1 and 1^3 + 1^2*(2^3 - 1^3) = 1+1*7 = 2^3, hence, k=1^3, cuban prime=7, and a(1)=m=2.
For n=3, q=A111251(3)=3 and (3^3)^3 + (3^3)^2*(4^3 - 3^3) = 27^3 + 27^2*37 = 46656 = 36^3, hence, k=3^3, cuban prime=37, and a(3)=m=36.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Project Euler, Problem 131: Prime cube partnership.

Crossrefs

Cf. A002407, A111251.

Subsequence of A011379.

Programs

Maple

for q from 1 to 90 do
p:=3*q^2+3*q+1;
if isprime(p) then print((q+1)*q^2); else fi; od:

Mathematica

f[n_] := n^2*(n+1); f /@ Select[Range[100], PrimeQ[3*#^2 + 3*# + 1] &] (* Amiram Eldar, Nov 05 2020 *)

PARI

lista(nn) =  apply(x->x^2*(x+1), select(x->isprime(3*x^2 + 3*x + 1), [1..nn])); \\ Michel Marcus, Nov 05 2020

Formula

a(n) = (A111251(n) + 1)*(A111251(n))^2.

a(n) = A011379(A111251(n)).